 I hope you understand when we are doing pore pressure parameters, we are talking about CU prime test and UU prime test, clear? Prime is the word with prime is how we differentiate them, consolidated undrained test with pore water pressure measurement, unconsolidated undrained test with pore water pressure measurement. So by means of a CU prime test, the shear strength parameters of the soil were found to be C prime equal to 10 kilo Newton per meter square, P prime equal to 24 degree. So this problem is the abridged version of the previous problem, clear? In the previous problem, we were shearing the sample and we reached up to failure, clear? And there we were computing A value, clear? All that has disappeared, the final state of the failure is known. So C prime, phi prime is known, your 50% job is done, you need not to understand the material, clear, under effective stress material, effective stress parameters. Now added to this is AF and what you are observing is a negative sign, fine? So what it corresponds to, what type of state of the soil is this? Very good. So it is a heavily over-consolidated material, clear? Heavily over-consolidated material will always show you suction, clear? So the pore water pressure which you are going to compute would be negative, that is it. So this is a type of characterization of the soil mass. Now in a UCS test, what UCS gives, remember? Go back to your lecture notes and check what UCS gives you, unconfined compression strength, what type of parameter gives you? So in a UCS test, a sample of the soil failed at a compressive stress of 162 kilo Newton per meter square, determine the initial value of suction in the soil. So this problem is a very straightforward direct problem, all right? No complications at all, everything is well known. Most of the time these type of tests are done, sorry, these type of analysis is done to check whether your pore water pressure measurement devices are working all right or not. You cannot believe, you have to calibrate them, clear? So I hope now you realize that from this what I am going to get is pore water pressure and pore water pressure I am measuring also, Cu prime, so consolidated undrained test with pore water pressure measurement. I want to check both the things. What could be another reason? Why I am so much eager to know about the pore water pressures? So this answers your question what you have been asking. Very soft, sensitive soils, the way you have made the sample, the way you have touched them and the way you have installed them on the pedestal might have changed the state of the materials also. In short, there are ways to check our experimentation and the results which we obtained from there. So rest of the problem is very simple. We have done this problem number 2 I suppose, where what we did is we obtained for C prime, 5 prime, the state of stress was given there and then we obtained pore water pressure if you remember. Now what we are doing is we are going to use the concept of AF over here. So AF is equal to delta U2 upon delta sigma 1-delta sigma 3. So what is given to you is a compressive stress of sigma 1-sigma 3 will be equal to 162 and because this is a UCS, so sigma 3 becomes 0. So sigma 1 becomes 162, how are you going to use the value of AF? So AF multiplied by 162 should be giving you delta U value alright. So what is the value of delta U? Delta U will be equal to initial pore water pressure minus the final pore water pressure and the pore water pressure which is coming from here. Let us say U2, rest is simple. So you have the Mohr-Coulomb envelope given to you, AF value was known, compute the suction value and this comes out to be 66.64. Be careful with the sign of the pore water pressure, it is a negative value in kilo Newton per meter square because if you compute the positive pore water pressure this will be redundant solution. What is the genesis of this pore water pressure development? Try to go by that alright. So you are achieving the failure by shearing the sample and then determine the initial value. This is a reverse problem of what we did just now, there we fixed that at 00 value something was given, I do not know that value. So I want to compute that. So what it says is determine the initial value of the suction in the soil sample before you started testing it in a triaxial test on a partially saturated soil. If you remember when we were talking about the capillarity in the soil and when we were computing gamma d values I had created a situation like this that you might be having a multi layered system of soils and somewhere here is the water table alright. Depending upon the type of the filling which you are going to use here and the compaction efforts, you are going to create a zone where saturation would be a function of Z. Everything below the water table is saturation equal to 1 here. We call this as a variably saturated soil deposit. Then I introduced the concept of capillary fringe here if you remember. So because of this water table over here and again depending upon the soil type and its compaction, you might be having a zone in which the capillarity dominates, is this part clear? We could compute this by using the hc value if you remember. So this thing we classified as hc equal to some 0.3 upon ed10 if you remember. This is an mm and this was in meters. So we have compute the capillary fringe also. What happens beyond capillary fringe is dry material unsaturated or variably saturated. So this becomes a very interesting problem. So the triaxial test on a partially saturated soil, it appears that I have taken a sample which is slightly above the capillary fringe. So it is not saturated, it is not submerged clear? Fine. It is a different state of the material which is defined like this. I was very eager to know now how the shear strength is changing, shear strength parameters are changing as a function of z because the question was that where should I lay the foundation? Correct? So I have given you a profile which I can obtain by doing investigations. I will do sampling, I will do different type of non-investing, non-invasive test, geophysical investigations. I will establish the subsurface profile clear? Now the question is where should I lay the foundations? So suppose if I decide that I will go for a shallow foundation or a strip footing clear? The question is what type of shear strength parameters are going to prevail there? This is the type of problem which we are doing. This is the background, okay? So there is a context for which you are doing triaxial testing, trying to obtain the parameters so that you can complete the design. So with all this story in mind at the back of your mind, now move ahead. So in a triaxial test on a partially saturated soil which is clay from particle size analysis from your Etterberg limits and all the characterization whatever you have studied until now, soil characterization clear? The sample was consolidated a sigma 3 of 150 kilo Newton per meters square after which the cell pressure was raised to 300 kilo Newton per meter square. The failure was achieved. I hope it is understood that you have sheared the sample. So DS is changing deviator stress and then you are failing the sample. Determine sigma 3 prime and sigma 1 prime at failure. First question is try to understand whether this problem can be solved or not without giving additional information. So quick answer is it cannot be because you need further information because failure condition is not defined yet properly. When we started this course, we were talking about failure. If you remember we said sigma tau at failure clear? And this was related to sigma x, sigma z, sigma y and tau series and angle of inclination if you remember at a point but the material did not come into the picture anywhere. That was the limitation of the Mohr's circle. Now once we go into the material, material can be depicted in several ways. Two easy ways would be either you talk about a parameter because a includes everything and we will see in the rest of the lecture today because a is a parameter which talks about the pore order pressures because of shearing related to OCR related to hydraulic conductivity related to particle size distribution clear and even plasticity index. So in short, if I want to solve this problem, I have to give the material properties. So if I give you C prime is pi prime is same as the previous problem, only thing is saturated sample, unsaturated sample less than 0.95 and AF prime is given as 0.25 rest of the things are same. Another thing is train your mind in such a manner that answers to all these problems should be not more than 3, 4 steps that part will come to you with the more realization. So this hardly has much to be done. You will be getting sigma 3 prime as 300 kilo Newton per meter square and sigma 1 prime will be equal to 120. Now where is my friend who was talking about sigma 3 greater than sigma 1 and sigma 1 greater than sigma 3. Are you realizing something? So in effective form what has happened? So your Mohr circle has shifted on the second quadrant, AF value is positive, read the problem in a tri-shell testing on a partially saturated soil, the sample was consolidated at this. From this state sigma 3 was increased to 300, clear. So your delta sigma 3 becomes 150. Now listen delta sigma 3 becomes 150, you know the B parameter, you can go ahead with all those things, AF prime is known. So you know delta sigma 1 is delta sigma 3, try to fit in that Mohr circle within this envelope, that is it, do both the ways numerically as well as graphically. Most of the infrastructure projects which are going on everywhere. One common mistake which people do it is that they do not talk about the gain in shear strength of the soil concepts. Now suppose if this is the formation of the soil and this soil is poor, soil is poor means either void ratios are very high, moisture content is very high, this is compressible, MV values are very high, CC values are extremely high, very close to unity let us say or more than unity, CCs are very high, void ratios are extremely high, alright. And velocity index is extremely high, now this is the delight of a geotechnical engineer because this becomes a difficult situation to train, we call them as problematic soils. General public call it as a problematic soil, but those who are technologists for them this problematic word gets changed by very good, excellent that is the spirit and not only the soils, this is the conditions, so what are the domains in which we are active these days, offshore engineering, challenging conditions, improving the soils in the marshy lands, waterlogged areas, organic clays and so on. This is something new which has to be learnt by all of you, our generation could do geomechanics with good formations, unfortunately now good formations are not available, so you have to go and encroach upon the challenging grounds, fine. Now what happens is, suppose if I consider this point O, what is the state of stress at this point, say sigma V prime and k times sigma V prime, you have studied effective stress principle, so now I am using effective stress terms, okay. Now starting from this state, if I load this system under undrained conditions, what is going to happen? This system will get converted to let us say sigma V prime plus let us say delta sigma clear and what is going to happen over here? This will also get changed to k sigma plus delta sigma or a function of delta sigma, I would try to tear as a multiplier of let us say A into delta sigma. The more and more I compress, what is happening? The system is getting improved, number one, that was the whole concept of one dimensional consolidation and by putting the, you know, pre-loading on the top of this deposit, so that the sample gets compressed, E values change, Cc decreases, Mb decreases and hence shear strength changes, so all these things are getting reflected into Cu. So truly speaking, what is happening is that Cu and delta sigma V prime, they play a balancing act, alright. So the more and more delta sigma V prime you apply, what is going to happen? Cu is going to enhance, why? Because this is the function of the material and this hypothesis was given by Schemton in 1957 that it is equal to 0.11 plus 0.0037 into Pi, this is what is known as Schemton's equation, 1957, ultimately this whole thing has to be analyzed. So if you follow the first principles, what you will be getting is the undrained cohesion or undrained shear strength of the material would be, do not remember this equation, my whole purpose of writing it over here is to demonstrate to you, so this is the form of the equation which you will be getting, importance of the parameters which we have talked about until now. Now K0 is the earth pressure coefficient at rest, AF is what we have discussed until now a parameter at failure. So what you are realizing is that there is a relationship between the undrained shear strength with the effective stresses which are getting caused on to the sample. Normally design charts are available between Cu upon delta sigma prime and they are plotted with respect to OCR and this is how you have these relationships, only thing you have to remember is for Cu equal to 0 is the condition for ANSI material, so truly speaking this function gets transformed to Cu upon delta sigma prime equal to this halter, so if this is 0 Cu upon delta sigma prime would be this parameter. One quick hint is if you do not include the change in the Cu value, there could be a situation where you might come across the consolidation which are more than 100% and that is not valid. There were beautiful bypasses which were designed in the country and the designer forgot this concept and to your surprise you will realize that the entire report the degree of consolidation achieved by more than 200, 300, 400% what this is known as is undrained shear strength and effective overburden relationship.